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AlgebraQuestion and Answers: Page 299

Question Number 70920    Answers: 1   Comments: 2

i ! = ?

$$\mathrm{i}\:!\:\:=\:\:? \\ $$

Question Number 70768    Answers: 1   Comments: 1

Question Number 70679    Answers: 0   Comments: 2

Hi, i wanna learn more english. can someone help me? write please +584249229498

$${Hi},\:{i}\:{wanna}\:{learn}\:{more}\:{english}. \\ $$$${can}\:{someone}\:{help}\:{me}? \\ $$$${write}\:{please}\:+\mathrm{584249229498} \\ $$

Question Number 70659    Answers: 1   Comments: 0

find the range algrbraically f(x)=(√(x^2 −1))

$${find}\:{the}\:{range}\:{algrbraically} \\ $$$$ \\ $$$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$

Question Number 70639    Answers: 0   Comments: 0

Question Number 70638    Answers: 1   Comments: 1

Question Number 70617    Answers: 1   Comments: 0

given that α and β are roots of the equation x^2 −5x + 4 =0 α>0 and β >0 find an equation whose roots are (√α) and (√β) how do i find (√(α )) + (√β)

$${given}\:{that}\:\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{2}} −\mathrm{5}{x}\:+\:\mathrm{4}\:=\mathrm{0}\: \\ $$$$\alpha>\mathrm{0}\:{and}\:\beta\:>\mathrm{0} \\ $$$${find}\:{an}\:{equation}\:{whose}\:{roots}\:{are}\:\sqrt{\alpha}\:{and}\:\sqrt{\beta}\: \\ $$$$ \\ $$$${how}\:{do}\:{i}\:{find}\:\:\sqrt{\alpha\:}\:+\:\sqrt{\beta}\: \\ $$

Question Number 70598    Answers: 1   Comments: 0

If a,b,c ∈ ℜ and (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b))=((12)/(a+b+c)) , (a/(b+c))+(b/(a+c))+(c/(a+b))=(4/3) then find a+b+c=?

$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\in\:\Re\:\mathrm{and}\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{12}}{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$$$,\:\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$

Question Number 70518    Answers: 2   Comments: 2

If a^3 −b^3 = 513 and ab= 54 then find a−b= ?

$$\mathrm{If}\:\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} =\:\mathrm{513}\:\mathrm{and}\:\mathrm{ab}=\:\mathrm{54}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\mathrm{a}−\mathrm{b}=\:? \\ $$

Question Number 70512    Answers: 1   Comments: 1

The faculty of science news paper reports that the combined membership of the mathematics club and science club is 122 students. What is the total membership of the chemistry club?

$$\mathrm{The}\:\mathrm{faculty}\:\mathrm{of}\:\mathrm{science}\:\mathrm{news}\:\mathrm{paper}\:\mathrm{reports}\:\mathrm{that}\:\mathrm{the}\:\mathrm{combined}\:\mathrm{membership}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{mathematics}\:\mathrm{club}\:\mathrm{and}\:\mathrm{science}\:\:\mathrm{club}\:\mathrm{is}\:\mathrm{122}\:\mathrm{students}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{membership}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chemistry}\:\mathrm{club}? \\ $$

Question Number 70460    Answers: 0   Comments: 2

Question Number 70312    Answers: 1   Comments: 1

If, (1/a^2 )+(1/b^2 )+(1/c^2 ) = (1/(ab))+(1/(bc))+(1/(ca)) then prove that, a=b=c.

$$\mathrm{If},\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}\:\mathrm{then}\:\mathrm{prove}\: \\ $$$$\mathrm{that},\:\mathrm{a}=\mathrm{b}=\mathrm{c}. \\ $$

Question Number 70216    Answers: 0   Comments: 0

let A = (((1 −1)),((1 2)) ) 1)calculate A^n 2) find e^A and e^(−A)

$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right){calculate}\:{A}^{{n}} \:\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\:{find}\:{e}^{{A}} \:{and}\:{e}^{−{A}} \\ $$

Question Number 70162    Answers: 1   Comments: 0

Question Number 70161    Answers: 1   Comments: 0

Question Number 70135    Answers: 0   Comments: 1

sophie−Germain identity a^4 +4b^4 =((a+b)^2 +b^2 )((a−b)^2 +b^2 )

$${sophie}−{Germain}\:{identity} \\ $$$${a}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{4}} =\left(\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}−{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$

Question Number 70103    Answers: 2   Comments: 0

if m^3 +2p^3 =3mn, a^3 +b^3 =p^3 and a^2 +b^2 =n then prove that a+b=m.

$$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$

Question Number 70040    Answers: 1   Comments: 3

If, a^2 b^2 c^2 ((1/a^3 )+(1/b^3 )+(1/c^3 ))=a^3 +b^3 +c^3 than prove that a,b,c Successive Proportional.

$$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$

Question Number 69995    Answers: 4   Comments: 0

1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))

$$\:\:\mathrm{1}.\begin{cases}{\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}^{+} \right] \\ $$$$ \\ $$$$\:\:\:\mathrm{2}.\begin{cases}{\sqrt{\sqrt{\mathrm{x}}+\mathrm{y}}+\sqrt{\mathrm{x}+\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{a}}}\\{\sqrt{\sqrt{\mathrm{x}}−\mathrm{y}}+\sqrt{\mathrm{x}−\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\mathrm{3}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right)}\end{cases} \\ $$$$ \\ $$$$\:\:\mathrm{4}.\begin{cases}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\boldsymbol{\mathrm{b}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} +\boldsymbol{\mathrm{z}}^{\mathrm{3}} =\boldsymbol{\mathrm{abxyz}}}\end{cases} \\ $$$$\:\:\:\mathrm{5}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\sqrt{\boldsymbol{\mathrm{x}}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\sqrt{\boldsymbol{\mathrm{x}}}}=\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{6}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}−\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$$$\:\mathrm{7}.\:\:\mathrm{16}^{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{16}^{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{10} \\ $$$$ \\ $$$$\:\:\mathrm{8}.\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{lnx}}} =\boldsymbol{\mathrm{x}}.\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\right) \\ $$

Question Number 69939    Answers: 2   Comments: 12

x,y,z ∈ Z^+ find all solutions of xy=(x+y)z

$${x},{y},{z}\:\in\:{Z}^{+} \\ $$$${find}\:{all}\:{solutions}\:{of}\:\boldsymbol{{xy}}=\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)\boldsymbol{{z}} \\ $$

Question Number 69741    Answers: 2   Comments: 4

Question Number 69738    Answers: 1   Comments: 0

Sarah dances everyday of the week, including saturdays and sundays. In november 2018, Sarah had to miss a few days. To control her absences she marks the day she missed class with a x on the calendar. She marked the 5th, 21st and 27th of november. What percentage indicates Sarah′s absences in november?

$${Sarah}\:{dances}\:{everyday}\:{of}\:{the}\:{week}, \\ $$$${including}\:{saturdays}\:{and}\:{sundays}. \\ $$$${In}\:{november}\:\mathrm{2018},\:{Sarah}\:{had}\:{to}\:{miss} \\ $$$${a}\:{few}\:{days}.\:{To}\:{control}\:{her}\:{absences} \\ $$$${she}\:{marks}\:{the}\:{day}\:{she}\:{missed}\:{class} \\ $$$${with}\:{a}\:\boldsymbol{{x}}\:{on}\:{the}\:{calendar}. \\ $$$${She}\:{marked}\:{the}\:\mathrm{5}{th},\:\mathrm{21}{st}\:{and}\:\mathrm{27}{th} \\ $$$${of}\:{november}. \\ $$$${What}\:{percentage}\:{indicates}\:{Sarah}'{s} \\ $$$${absences}\:{in}\:{november}? \\ $$

Question Number 71696    Answers: 1   Comments: 0

The side of a square is measured to be 12cm long cofrect to the nearest cm. Find the maximum absolute error and the maximum percentage error for (a) The length of the square (Answer: 0.5cm, 4.17%) (b) The area of the square. (Answer: 12.25cm, 8.5%)

$$\mathrm{The}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{12cm}\:\mathrm{long}\:\mathrm{cofrect} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\:\mathrm{cm}.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{absolute}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{percentage}\:\mathrm{error}\:\mathrm{for} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\:\:\:\:\left(\mathrm{Answer}:\:\:\mathrm{0}.\mathrm{5cm},\:\:\mathrm{4}.\mathrm{17\%}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}.\:\:\:\:\:\:\left(\mathrm{Answer}:\:\:\:\:\mathrm{12}.\mathrm{25cm},\:\:\:\mathrm{8}.\mathrm{5\%}\right) \\ $$

Question Number 69681    Answers: 2   Comments: 2

Question Number 69665    Answers: 1   Comments: 1

Question Number 69662    Answers: 1   Comments: 0

prove that ((2x^3 −x^2 −2x+1)/(x^3 +1)) + ((x^3 +1)/(x^4 −2x^3 +3x^2 −2x+1)) = 2

$${prove}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\:=\:\mathrm{2} \\ $$

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